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Math Help - Chance of pairs, rolling 5 dices simultaneously

  1. #1
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    Unhappy Chance of pairs, rolling 5 dices simultaneously

    Hello all,

    Can you please help me solve the following problem?
    This is from A first course in probability by Seldon Ross 7th edition.

    Chapter 2 Problem 16
    Poker dice is played by simultaneously rolling 5 dice. Show that:
    a) P { no two alike } = 0.0926
    b) P { one pair } = 0.4630
    c) P { two pair } = 0.2315
    d) P { three alike } = 0.1543
    e) P { full house } = 0.0386
    f) P { four alike } = 0.0193
    g) P { five alike } = 0.0008

    Thanks!
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  2. #2
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    Ok I can solve problems a,b and g.

    a) P { no two alike } = \frac{6*5*4*3*2}{6^5} = 0.0926


    b) P { one pair } = \frac{6*{5 \choose 2}*5*4*3}{6^5} = 0.4630


    g) P { five alike } = \frac{6}{6^5} = 0.0008

    However the trick I used for b doesn't work for c,d,e and f, and I don't understand
    why? This is how I solve b:

    The numerical value for the pair can be chosen in 6 ways and can be placed in {5 \choose 2} ways, and then there are 3 'slots' left. The numerical value for the remaining 3 slots can be chosen in 5*4*3 ways. The whole thing is divided by the total number of possibilities which is 6^5.

    But as I said this same line of reasoning doesn't work for c,d,e and f? Am I doing something wrong?
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  3. #3
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    Quote Originally Posted by desperate87 View Post
    Poker dice is played by simultaneously rolling 5 dice. Show that:
    c) P { two pair } = 0.2315
    d) P { three alike } = 0.1543
    e) P { full house } = 0.0386
    f) P { four alike } = 0.0193
    It is just counting. All the dominators 6^5.

    c) Pick the two pairs: {6 \choose 2}. Pick the odd die out 4. Now scramble.
    {6 \choose 2}4\frac{5!}{(2!)^2}

    You try the rest!
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  4. #4
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    Thank you Plato
    I finally understand it.

    I'll post my results when I'm done
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